I have this String:
"[" & vbCrLf & " ""APPLE""" & vbCrLf & "]"
The only thing I need is APPLE.
I tried a few options with Split, Trim, Left and more, but they didn't work very well.
Thank you very much!
As the comments above have said, there's not enough information to give an answer without making assumptions, which could be wrong. I've assumed you want to extract the value between two quotation marks, regardless of what else is before or after.
If that's what you want, try this:
Dim Result As String = Nothing
Dim source As String = $"[{vbCrLf}""Apple""{vbCrLf}]"
Dim FirstQuote As Integer = source.IndexOf("""")
If FirstQuote > -1 And source.Length > FirstQuote Then
Dim SecondQuote As Integer = source.IndexOf("""", FirstQuote + 1)
If SecondQuote > FirstQuote Then
Result = source.Substring(FirstQuote + 1, SecondQuote - FirstQuote - 1)
End If
End If
If Result Is Nothing Then
'Handle Invalid Format
Else
'Process Result
End If
You would need to modify that so that you passed your source string, rather than defining it in the code. If you wanted to extract multiple words from a single string in the same format, just set FirstQuote = SecondQuote + 1, check that doesn't exceed the length of the source string and loop through again.
I am going to assume that you probably just need to get the first occurance of a string (in this case "apple") within square-brackets using split and so:
Dim AppleString As String = "This is an [Apple] or etc [...]"
console.WriteLine(AppleString.split("[")(1).split("]")(0).trim())
⚠️ This is not a solution for all purposes !!!
Related
I'm a programing student, so I've started with vb.net as my first language and I need some help.
I need to know how I delete excess white spaces between words in a sentence, only using these string functions: Trim, instr, char, mid, val and len.
I made a part of the code but it doesn't work, Thanks.
enter image description here
Knocked up a quick routine for you.
Public Function RemoveMyExcessSpaces(str As String) As String
Dim r As String = ""
If str IsNot Nothing AndAlso Len(str) > 0 Then
Dim spacefound As Boolean = False
For i As Integer = 1 To Len(str)
If Mid(str, i, 1) = " " Then
If Not spacefound Then
spacefound = True
End If
Else
If spacefound Then
spacefound = False
r += " "
End If
r += Mid(str, i, 1)
End If
Next
End If
Return r
End Function
I think it meets your criteria.
Hope that helps.
Unless using those VB6 methods is a requirement, here's a one-line solution:
TextBox2.Text = String.Join(" ", TextBox1.Text.Split(New Char() {" "c}, StringSplitOptions.RemoveEmptyEntries))
Online test: http://ideone.com/gBbi55
String.Split() splits a string on a specific character or substring (in this case a space) and creates an array of the string parts in-between. I.e: "Hello There" -> {"Hello", "There"}
StringSplitOptions.RemoveEmptyEntries removes any empty strings from the resulting split array. Double spaces will create empty strings when split, thus you'll get rid of them using this option.
String.Join() will create a string from an array and separate each array entry with the specified string (in this case a single space).
There is a very simple answer to this question, there is a string method that allows you to remove those "White Spaces" within a string.
Dim text_with_white_spaces as string = "Hey There!"
Dim text_without_white_spaces as string = text_with_white_spaces.Replace(" ", "")
'text_without_white_spaces should be equal to "HeyThere!"
Hope it helped!
I have a difficult situation and so far no luck in finding a solution.
My VBA collects number figures like $80,000.50. and I'm trying to get VBA to remove the last period to make it look like $80,000.50 but without using right().
The problem is after the last period there are hidden spaces or characters which will be a whole lot of new issue to handle so I'm just looking for something like:
replace("$80,000.50.",".**.",".**")
Is this possible in VBA?
I cant leave a comment so....
what about InStrRev?
Private Sub this()
Dim this As String
this = "$80,000.50."
this = Left(this, InStrRev(this, ".") - 1)
Debug.Print ; this
End Sub
Mid + Find
You can use Mid and Find functions. Like so:
The Find will find the first dot . character. If all the values you are collecting are currency with 2 decimals, stored as text, this will work well.
The formula is: =MID(A2,1,FIND(".",A2)+2)
VBA solution
Function getStringToFirstOccurence(inputUser As String, FindWhat As String) As String
getStringToFirstOccurence = Mid(inputUser, 1, WorksheetFunction.Find(FindWhat, inputUser) + 2)
End Function
Other possible solutions, hints
Trim + Clear + Substitute(Char(160)): Chandoo -
Untrimmable Spaces – Excel Formula
Ultimately, you can implement Regular expressions into Excel UDF: VBScript’s Regular Expression Support
How about:
Sub dural()
Dim r As Range
For Each r In Selection
s = r.Text
l = Len(s)
For i = l To 1 Step -1
If Mid(s, i, 1) = "." Then
r.Value = Mid(s, 1, i - 1) & Mid(s, i + 1)
Exit For
End If
Next i
Next r
End Sub
This will remove the last period and leave all the other characters intact. Before:
and after:
EDIT#1:
This version does not require looping over the characters in the cell:
Sub qwerty()
Dim r As Range
For Each r In Selection
If InStr(r.Value, ".") > 0 Then r.Characters(InStrRev(r.Text, "."), 1).Delete
Next r
End Sub
Shortest Solution
Simply use the Val command. I assume this is meant to be a numerical figure anyway? Get rid of commas and the dollar sign, then convert to value, which will ignore the second point and any other trailing characters! Robustness not tested, but seems to work...
Dim myString as String
myString = "$80,000.50. junk characters "
' Remove commas and dollar signs, then convert to value.
Dim myVal as Double
myVal = Val(Replace(Replace(myString,"$",""),",",""))
' >> myVal = 80000.5
' If you're really set on getting a formatted string back, use Format:
myString = Format(myVal, "$000,000.00")
' >> myString = $80,000.50
From the Documentation,
The Val function stops reading the string at the first character it can't recognize as part of a number. Symbols and characters that are often considered parts of numeric values, such as dollar signs and commas, are not recognized.
This is why we must first remove the dollar sign, and why it ignores all the junk after the second dot, or for that matter anything non numerical at the end!
Working with Strings
Edit: I wrote this solution first but now think the above method is more comprehensive and shorter - left here for completeness.
Trim() removes whitespace at the end of a string. Then you could simply use Left() to get rid of the last point...
' String with trailing spaces and a final dot
Dim myString as String
myString = "$80,000.50. "
' Get rid of whitespace at end
myString = Trim(myString)
' Might as well check if there is a final dot before removing it
If Right(myString, 1) = "." Then
myString = Left(myString, Len(myString) - 1)
End If
' >> myString = "$80,000.50"
I am writing VBA code for an Excel workbook. I would like to be able to open a connection with an Access database, and then import a txt file (pipe delimited) and create a new table in the database from this txt file. I have searched everywhere but to no avail. I have only been able to find VBA code that will accomplish this from within Access itself, rather than from Excel. Please help! Thank you
Google "Open access database from excel VBA" and you'll find lots of resources. Here's the general idea though:
Dim db As Access.Application
Public Sub OpenDB()
Set db = New Access.Application
db.OpenCurrentDatabase "C:\My Documents\db2.mdb"
db.Application.Visible = True
End Sub
You can also use a data access technology like ODBC or ADODB. I'd look into those if you're planning more extensive functionality. Good luck!
I had to do this exact same problem. You have a large problem presented in a small question here, but here is my solution to the hardest hurdle. You first parse each line of the text file into an array:
Function ParseLineEntry(LineEntry As String) As Variant
'Take a text file string and parse it into individual elements in an array.
Dim NumFields As Integer, LastFieldStart As Integer
Dim LineFieldArray() As Variant
Dim i As Long, j As Long
'Determine how many delimitations there are. My data always had the format
'data1|data2|data3|...|dataN|, so there was always at least one field.
NumFields = 0
For I = 1 To Len(LineEntry)
If Mid(LineEntry, i, 1) = "|" Then NumFields = NumFields + 1
Next i
ReDim LineFieldArray(1 To NumFields)
'Parse out each element from the string and assign it into the appropriate array value
LastFieldStart = 1
For i = 1 to NumFields
For j = LastFieldStart To Len(LineEntry)
If Mid(LineEntry, j , 1) = "|" Then
LineFieldArray(i) = Mid(LineEntry, LastFieldStart, j - LastFieldStart)
LastFieldStart = j + 1
Exit For
End If
Next j
Next i
ParseLineEntry = LineFieldArray
End Function
You then use another routine to add the connection in (I am using ADODB). My format for entries was TableName|Field1Value|Field2Value|...|FieldNValue|:
Dim InsertDataCommand as String
'LineArray = array populated by ParseLineEntry
InsertDataCommand = "INSERT INTO " & LineArray(1) & " VALUES ("
For i = 2 To UBound(LineArray)
If i = UBound(LineArray) Then
InsertDataCommand = InsertDataCommand & "'" & LineArray(i) & "'" & ")"
Else
InsertDataCommand = InsertDataCommand & LineArray(i) & ", "
End If
Next i
Just keep in mind that you will have to build some case handling into this. For example, if you have an empty value (e.g. Val1|Val2||Val4) and it is a string, you can enter "" which will already be in the ParseLineEntry array. However, if you are entering this into a number column it will fail on you, you have to insert "Null" instead inside the string. Also, if you are adding any strings with an apostrophe, you will have to change it to a ''. In sum, I had to go through my lines character by character to find these issues, but the concept is demonstrated.
I built the table programmatically too using the same parsing function, but of this .csv format: TableName|Field1Name|Field1Type|Field1Size|...|.
Again, this is a big problem you are tackling, but I hope this answer helps you with the less straight forward parts.
Hello I am trying to get part of my string out and was wondering if anyone can help so here is what I am working with
Dim tempName, NewName As String
'This is an example of what tempName will Equal
'What I need is the information in between the first -
'and the second one. In this case it would be 120
'I can not do it by number because the dashes are the only thing
'that stays the same.
tempName = "3-120-12-6"
NewName = tempName 'Do not know what String Manipulation to use.
Another few examples would be 6-56.5-12-12 I need the 56.5 or 2-89-12-4 I would need the 89
Thank you for the help.
You can do this too... You can break it up into pieces by detecting the first -. Then get the second...
Dim x as String = Mid(tempName, InStr(tempName, "-") + 1)
NewName = Mid(x, 1, InStr(x , "-") - 1)
Or make something like this to get it into 1 line of code...
NewName = Mid(Mid(tempName, InStr(tempName, "-") + 1), 1, InStr(Mid(tempName, InStr(tempName, "-") + 1), "-") - 1)
Or the quickest approach to this would be to use Split like the code below... This code equates the values between - into an array. Since you want the second value, you'd want the array with the index of 1 since an array starts with 0. If you're only starting, do try to create your own way to manipulate the string, this will help your mind think of new things and not just rely on built-in functions...
Dim tempName As String = "3-120-12-6"
Dim secondname() As String = Split(tempName, "-")
NewName = secondname(1)
I have a powershell script that pulls out lines containing ".html" ".css" and so forth
however what I need is to be able to strip out the entire filename
using a pattern.... the entire pattern is returned example
.........\.html returns
src="blank.html"
my answer came in VB (with a bunch of work and even more research) I wanted to share with you all the results, it's not pretty but it works. is there an easier way?
I have commented the code to help in understanding.
Private Sub find()
Dim reader As StreamReader = My.Computer.FileSystem.OpenTextFileReader(openWork.FileName)
Dim a As String
Dim SearchForThis As String
Dim allfilenames As New System.Text.StringBuilder
Dim first1 As String
Dim FirstCharacter As Integer
'Dim lines As Integer
SearchForThis = txtFind.Text
Do
a = reader.ReadLine 'reader.Readling
If a = "" Then
a = reader.ReadLine
End If
If a Is Nothing Then 'without this check the for loops run with bad data, but I can't check "a" without reading it first.
Else
For FirstCharacter = 2 To a.Length - SearchForThis.Length ' start at 2 to prevent errors in the ")" check
If Mid(a, FirstCharacter, SearchForThis.Length) = SearchForThis Then ' compare the line character by character to find the searchstring
If Mid(a, FirstCharacter - 1, 1) <> ")" Then ' checks for ")" just before the searchstring (a common problem with my .CSS finds)
For y = FirstCharacter To 1 Step -1
If Mid(a, y, 1) = Mid(a, FirstCharacter + SearchForThis.Length, 1) Then ' compares the character after searchstring till I find another one
Dim temp = Mid(a, y + 1, (FirstCharacter + SearchForThis.Length) - 1 - y) ' puts the entire filename into variable "temp"
allfilenames.Append(temp & Chr(13)) 'adds the contents of temp (and a carrage return) to the allfilenames stringbuilder
y = 1
Else
End If
Next
End If
End If
Next
End If
Loop Until a Is Nothing
Document.Text = allfilenames.ToString
reader.Close()
End Sub
(updating for comments... thanks for the input)
each line in the .css search file looks something like this.
addPbrDlg.html:12:<link rel=stylesheet href="swl_styles-5.0o-97581885.css" TYPE="text/css">
addPbrDlg.html:727: html(getFrame(statusFrame).strErrorMessage).css('color','red');
for this I want to return
swl_styles-5.0o-97581885.css
but not return
statusFrame).strErrorMessage).css
basically I want to strip out the file names from HTML code
but if I use a pattern like
.............................\.css
it would return something like
t href="swl_styles-5.0o-97581885.css
Finally... there are some variables that I don't need to worry about (due to my personal situation) like I know that all web pages are ".html" all images are ".gif" there are ".css" and ".js" files as well that I want to pull. But because the designers are extremely consistant I know that there aren't any surprise files (.jpg or .htm)
I can also assume that if there is a single quote after the filename, there will be a single quote before. same with double quote.
Thanks for your input so far... I appreciate your time and knowledge.
You need to use Regex and do something like this
Dim files = Regex.Matches("<your whole file text>", "Your regex pattern");
Your regex pattern will look something like this "\Asrc="".+((\.html)|(\.css))"")". This is probably wrong but when you get that straight follow with
Dim fileList as new List(of String)
For Each file as Match in files
' strip " src=" " and last " " "
fileList.Add(file.Value.Substring(5, file.Value.Length - 6))
Next